1. If you had used 0.030 g of magnesium in a first trial and 0.040 g in a second
ID: 995031 • Letter: 1
Question
1. If you had used 0.030 g of magnesium in a first trial and 0.040 g in a second, would you expect the molar volume at STP to be larger or smaller in the second trial? Explain your answer. Do not merely explain by saying molar volume is an intensive property. Explain WHY its molar volume is not dependent on the sample size.
2. Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)
a) In order for this reaction to work properly to give the correct molar volume of an ideal gas, which must be the limiting reactant in the reaction? Explain your answer.
b). If there is an unreacted metal remaining at the end of the reaction, what would that do to your calculated molar volume? Be specific. Do not merely say that “it would be
Explanation / Answer
1) Molar volume (Vm) would not change with size in Magnesium metal case.
Molar volume is defined as the volume occupied by the given compound.
Formula,
Density = Mass/Volume
Volume = Mass/Density
Hence,
Molar volume = Molar mass /Density
At Standard temperature and pressure conditions (T = 273.15 K and P = 1 atm) Magnesium metal exist in solid state.
For solid actual volume occupied is negligibly small and hence with increase in mass i.e. size molar volume changes only negligibly.
Hence in solid case molar volume is an intensive property i.e. molar volume is independent of size of the solid sample.
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2) Mg (s) + 2HCl ---------> MgCl2 (aq.) + H2 (g)
Molar volume of Mg metal is constant.
HCl is liquid at STP condition hence molar volume may subject to change.
Hence it will be better if we use Mg as limiting reagent which will hence quantitatively react with HCl and gives formation of MgCl2 and H2 gas.
Quantitative relationship between Mg metal and H2 gas (Whose ideality under question) is,
1 mole of Mg = 1 mole H2 gas
40 g of Mg = 2 g of H2 gas
We can way Mg metal before it react with HCl to form H2 gas
And from the mass of Mg used we can calculate mass of H2 gas produced and also the volume of H2 gas at STP can be calculated using Avogadro’s law states that at STP volume occupied by 1 mole of (i.e. 2 g in H2 gas case) gas is 22.4 L.
Hence the molar volume calculation.
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3)
If some amount of Mg left unreacted amount of H2 gas formed will be less than expected.
In calculation we will use mass of H2 g formed equivalent to mass of Mg.
But as actual mass of Mg reacted is less than calculated it will give wrongly higher value of Mass of H2 and hence the faulty higher value of molar volume of H2 gas.
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